Problem: Define a sphere: $S = \{ (x, y, z) \in \mathbb{R}^3 \big | x^2 + y^2 + z^2 \leq 6 \}$ What is the triple integral of the scalar field $f(x, y, z) = xy - z$ over $S$ in spherical coordinates? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^\pi \int_0^{2\pi} \int_{-3}^3 \rho^4 \sin(\theta) \sin^3(\varphi) - \rho^3 \sin(\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice B) B $\dfrac{1}{2} \int_0^\pi \int_0^{2\pi} \int_0^6 \rho^4 \sin(2\theta) \sin^3(\varphi) - \rho^3 \sin(2\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice C) C $ \int_0^\pi \int_0^{2\pi} \int_0^6 \rho^2 \sin(\theta) \sin^3(\varphi) - \rho \sin(\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice D) D $\dfrac{1}{2} \int_0^\pi \int_0^{2\pi} \int_{-3}^3 \rho^2 \sin(2\theta) \sin^3(\varphi) - \rho \sin(2\varphi) \, d\rho \, d\theta \, d\varphi$
Explanation: The bounds we'll use are $0 < \theta < 2\pi$ and $0 < \varphi < \pi$. Here is the change of variables for spherical coordinates. $\begin{aligned} x &= \rho \cos(\theta) \sin(\varphi) \\ \\ y &= \rho \sin(\theta) \sin(\varphi) \\ \\ z &= \rho \cos(\varphi) \end{aligned}$ We want to represent the sphere $S$ with bounds in spherical coordinates. The standard unit sphere needs $\varphi$ to range across $[0, \pi]$, $\theta$ to range across $[0, 2\pi]$, and $\rho$ to range across $[0, 1]$. The sphere $S$ has radius $6$, so it needs $\rho$ to range across $[0, 6]$. $ \int_0^\pi \int_0^{2\pi} \int_0^6 \cdots \, d\rho \, d\theta \, d\varphi$ We can now put $f(x, y, z)$ in the integrand, but we need to substitute $x$, $y$, and $z$ for their definitions in spherical coordinates. $\begin{aligned} & \int_0^\pi \int_0^{2\pi} \int_0^6 (xy - z) \cdots \, d\rho \, d\theta \, d\varphi \\ \\ &= \int_0^\pi \int_0^{2\pi} \int_0^6 (\rho^2 \cos(\theta)\sin(\theta) \sin^2(\varphi) - \rho\cos(\varphi)) \cdots \, d\rho \, d\theta \, d\varphi \end{aligned}$ The final step is finding the Jacobian of spherical coordinates, which we'll need to multiply in to get the final integral. $J(\rho, \theta, \varphi) = \rho^2\sin(\varphi)$ [Derivation] We can simplify the product using the double angle identity that $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$. $\begin{aligned} & \int_0^\pi \int_0^{2\pi} \int_0^6 (\rho^2 \cos(\theta)\sin(\theta) \sin^2(\varphi) - \rho\cos(\varphi)) \rho^2\sin(\varphi) \, d\rho \, d\theta \, d\varphi \\ \\ &= \dfrac{1}{2} \int_0^\pi \int_0^{2\pi} \int_0^6 \rho^4 \sin(2\theta) \sin^3(\varphi) - \rho^3 \sin(2\varphi) \, d\rho \, d\theta \, d\varphi \end{aligned}$ The integral in spherical coordinates: $\dfrac{1}{2} \int_0^\pi \int_0^{2\pi} \int_0^6 \rho^4 \sin(2\theta) \sin^3(\varphi) - \rho^3 \sin(2\varphi) \, d\rho \, d\theta \, d\varphi$